# Prove $\left|\sin\frac1x\right| \leq 1$, $\left| x\sin\frac1x\right| \leq |x|$ for all $x \neq 0$

I saw the following statement in a textbook:

Since $$\left|\sin\frac1x\right| \leq 1$$, $$\left| x\sin\frac1x\right| \leq |x|$$ for all $$x \neq 0$$.

The author didn't show show they come to the conclusion, I tried to prove it myself and got confused:

$$\left|\sin\frac1x\right| \leq 1 \implies -1\leq\sin\frac1x\ \leq 1 \implies -|x|\leq|x|\sin\frac1x\ \leq |x| \implies \left| |x|\sin\frac1x\right|\leq |x|$$

Now I need to prove that

$$\left|x\sin\frac1x\right|\leq \left| |x|\sin\frac1x\right|$$ to draw the same conclusion where I've got overwhelmed by the number of cases and wasn't sure if that was the way to prove the original implication. Intuitively it's clear that the left and right parts of the inequality are equal. What a formal proof would look like?

• $|a| |b| = |ab|$ Oct 24, 2020 at 13:51

## 2 Answers

The author simply used the fact that$$\left|x\sin\left(\frac1x\right)\right|=|x|\left|\sin\left(\frac1x\right)\right|\leqslant|x|.$$

$$|x\sin \frac{1}{x} | = |x| |\sin \frac{1}{x} | \leq |x|$$